a) 4.x - 15 = -75 - x

=> 4x + x = -75 + 15

=> 5x = -60

=> x = -60/5 = -12

b) 72 - 3.x = 5.x + 8

=> -3x - 5x = 8 - 72

=> -8x = -64

=> x = -64/-8 = 8

                                                           Bài giải

a, \(4\cdot x-15=-75-x\)

\(4x+x=-75+15\)

\(5x=-60\)

\(x=-60\text{ : }5\)

\(x=-12\)

b, \(72-3\cdot x=5\cdot x+8\)

\(5x+3x=72-8\)

\(8x=64\)

\(x=64\text{ : }8\)

\(x=8\)

a, \(4\cdot x-15=-75-x\)

\(4x+x=-75+15\)

\(5x=-60\)

\(x=-60\text{ : }5\)

\(x=-12\)

b, \(72-3\cdot x=5\cdot x+8\)

\(5x+3x=72-8\)

\(8x=64\)

\(x=64\text{ : }8\)

\(x=8\)

Y x 4 + y x 3 + y = 720

Y x 4 + y x 3 + y x 1 = 720

y x (4 + 3 + 1) = 720
y x 8 = 720

y = 720 : 8

y = 90

\(a,5x^2-10xz+xy-2yz\\ =5x\left(x-2z\right)+y\left(x-2z\right)\\ =\left(5x+y\right)\left(x-2z\right)\\ b,9x^2-3x-y^2+y\\ =\left(3x-y\right)\left(3x+y\right)-\left(3x-y\right)\\ =\left(3x-y\right)\left(3x+y-1\right)\\ c,y^2-z^2+12z-36\\ =y^2-\left(z-6\right)^2\\ =\left(y-z+6\right)\left(y+z-6\right)\\ d,2y^2-8z^2+\left(y-2z\right)^3\\ =2\left(y-2z\right)\left(y+2z\right)+\left(y-2z\right)^3\\ =\left(y-2z\right)\left(y^2-4yz+4z^2+2y+4z\right)\)

cứu giúp 

\(a,5x^2-10xz+xy-2yz\\ =5x\left(x-2z\right)+y\left(x-2z\right)\\ =\left(5x+y\right)\left(x-2z\right)\\ b,9x^2-3x-y^2+y\\ =\left(9x^2-y^2\right)-\left(3x-y\right)\\ =\left(3x-y\right)\left(3x+y\right)-\left(3x-y\right)\\ =\left(3x-y\right)\left(3x+y-1\right)\)

\(-\left|1,7-x\right|-\dfrac{5}{3}=\dfrac{2}{3}\\ \Rightarrow\left|1,7-x\right|=-\dfrac{5}{3}-\dfrac{2}{3}=-\dfrac{7}{3}\left(l\right)\)

Vậy không có giá trị x thoả mãn

\(\left(x+y\right)^3-\left(x-y\right)^3=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3=6x^2y+2y^3\)

\(\left(x-3\right)\cdot\left(y-5\right)=3\)

=>\(\left(x-3\right)\cdot\left(y-5\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)

=>\(\left(x-3;y-5\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(4;8\right);\left(6;6\right);\left(2;2\right);\left(0;4\right)\right\}\)

=>5căn x+2-15y=15 và 5căn x+2-2y=71/3

=>-13y=4/3 và căn x+2-3y=3

=>y=-4/39 và căn x+2=3+3y=3-12/39=105/39

=>y=-4/39 và x=887/169

=>2x=2/3

hay x=1/3

\(\dfrac{9}{7}\times x+\dfrac{5}{7}\times x=\dfrac{2}{3}\)

\(\dfrac{14}{7}\times x=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:2=\dfrac{1}{3}\)

2x=2/3

hay x=1/3

27:(x-3/2)^3=(x-3/2):3

Ta có: \(\dfrac{27}{\left(x-\dfrac{3}{2}\right)^3}=\dfrac{\left(x-\dfrac{3}{2}\right)}{3}\)

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^3.\left(x-\dfrac{3}{2}\right)\)=27.3

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4\)=81

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4=3^4\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}+\dfrac{3}{2}\\x=\dfrac{-8}{2}+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)

Vậy x∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)